Solve for $x$ : $3x^2 + 3x - 18 = 0$
Answer: Dividing both sides by $3$ gives: $ x^2 + {1}x {-6} = 0 $ The coefficient on the $x$ term is $1$ and the constant term is $-6$ , so we need to find two numbers that add up to $1$ and multiply to $-6$ The two numbers $3$ and $-2$ satisfy both conditions: $ {3} + {-2} = {1} $ $ {3} \times {-2} = {-6} $ $(x + {3}) (x {-2}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 3) (x -2) = 0$ $x + 3 = 0$ or $x - 2 = 0$ Thus, $x = -3$ and $x = 2$ are the solutions.